Problem: $\dfrac{ 2n + 5p }{ 10 } = \dfrac{ -n + 3q }{ -7 }$ Solve for $n$.
Answer: Multiply both sides by the left denominator. $\dfrac{ 2n + 5p }{ {10} } = \dfrac{ -n + 3q }{ -7 }$ ${10} \cdot \dfrac{ 2n + 5p }{ {10} } = {10} \cdot \dfrac{ -n + 3q }{ -7 }$ $2n + 5p = {10} \cdot \dfrac { -n + 3q }{ -7 }$ Multiply both sides by the right denominator. $2n + 5p = 10 \cdot \dfrac{ -n + 3q }{ -{7} }$ $-{7} \cdot \left( 2n + 5p \right) = -{7} \cdot 10 \cdot \dfrac{ -n + 3q }{ -{7} }$ $-{7} \cdot \left( 2n + 5p \right) = 10 \cdot \left( -n + 3q \right)$ Distribute both sides $-{7} \cdot \left( 2n + 5p \right) = {10} \cdot \left( -n + 3q \right)$ $-{14}n - {35}p = -{10}n + {30}q$ Combine $n$ terms on the left. $-{14n} - 35p = -{10n} + 30q$ $-{4n} - 35p = 30q$ Move the $p$ term to the right. $-4n - {35p} = 30q$ $-4n = 30q + {35p}$ Isolate $n$ by dividing both sides by its coefficient. $-{4}n = 30q + 35p$ $n = \dfrac{ 30q + 35p }{ -{4} }$ Swap signs so the denominator isn't negative. $n = \dfrac{ -{30}q - {35}p }{ {4} }$